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The Question

Solve the equation in natural numbers:
$x^3+2x+1=y^2,x,y\in N$
The solution is:
$x=1,y=2;x=8,y=23$

My Understanding

The given equation can be written as $x(x^2 +2)=(y-1)(y+1)$, as we're working in $\mathbb{N}$ we've the following cases:

  • Case 1: $x=y-1$ and $x^2 +2 =y+1$ using these we get $y^2 -3y +2=(y-2)(y-1)=0$ therfore it follows that $(x,y)=(1,2)$ is the only solution in this case.
  • Case 2: $x=y+1$, and $x^2+2 =y-1$ , using these we get $y^2 +y+4=0$. As the quadratic is irreducible over $\mathbb{Z}$ we can conclude the non existence of solutions in this case.

Therefore we can conclude that the only solution in $\mathbb{N}^{2}$ is $(x,y)=(1,2)$

What about $(8,23)$?

nonuser
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Ozod
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1 Answers1

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Solving this problem using Elliptic Curves is probably not the best idea.

Examining the solutions to $E:y^2=x^3+2x+1$ over finite fields we see that $E(\mathbb{F}_{11})=16$ and $E(\mathbb{F}_{19})=27$. This mean that $|E(\mathbb{Q})_{\mathrm{tors}}|$ divides $16\cdot11^{r_1}$ for some $r_1$ and $|E(\mathbb{Q})_{\mathrm{tors}}|$ divides $27\cdot19^{r_2}$ for some $r_2$. The only solution is $|E(\mathbb{Q})_{\mathrm{tors}}|=1$, and so the torsion group is trivial.

As for the rank of the curve, the method of descent applied with $p=2$ shows that $E$ has rank $1$ (this can be done in Magma or Sage).

Combining both results, we see that $E(\mathbb{Q})$ is generated entirely by the free point $(0,1)$ on the curve. The issue is, there is no simple way to reduce this to just the solutions over $\mathbb{Z}$. It defiantly can be done, Magma or Sage can tell us that the the points $(0,\pm1), (1,\pm 2)$ and $(8,\pm 23)$ are the only solutions. If this question is being asked to high schoolers or undergraduates, this is not the way to go about it.

Milo Moses
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    You don't need to work as hard as the L-function. One can simply perform a $2$-descent on $E$ to determine that the $2$-Selmer group has rank $1$ (as an $\mathbb{F}_2$ module). This is implemented in e.g., Magma as TwoDescent(E) - but in this case it can be done by hand. – Mummy the turkey Feb 14 '21 at 19:53
  • Yes, that would have been simpler :) – Milo Moses Feb 14 '21 at 19:54
  • Additionally one does not need envoke Mazur for the torsion subgroup, $E(\mathbb{Q})[n]$ injects into either $E(\mathbb{F}{11})$ or $E(\mathbb{F}{19})$ (see e.g., Silverman VII 3.1) an we're done (with this in mind you can just check mod 3 and mod 7) – Mummy the turkey Feb 14 '21 at 20:07
  • $E(\mathbb{Q})[n]$ injects into $E(\mathbb{F}{11})$ and $E(\mathbb{F}{13})$ only if $11$ or $13$ do not divide $n$. Mazur's theorem needs to be invoked to cover the case where $n=11$ or $n=13$ – Milo Moses Feb 14 '21 at 20:27
  • @Milo Moses I corrected $(8,\pm27)$ to $(8,\pm23).\space$ I hope you don't mind. – poetasis Feb 14 '21 at 20:33
  • @MiloMoses From the first you have $#E(\mathbb{Q}){tors}$ divides $11^r#E(\mathbb{F}{11}) = 16\cdot 11^r$ for some $r$. From the second you have $#E(\mathbb{Q}){tors}$ divides $19^s #E(\mathbb{F}{19}) = 27 \cdot 19^s$ for some $s$. Hence you can immediately conclude. No need for Mazur – Mummy the turkey Feb 14 '21 at 20:46
  • Ah yes, I see your argument now. I will rewrite my answer to this shortened form – Milo Moses Feb 14 '21 at 22:59
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    No worries, it's good to not invoke sledgehammers when the methods one can use here are pretty old. I'd also mention that you say "not for the faint of heart" - finiteness of $\mathbb{Z}$ points is garenteed, and there are effective theorems, c.f., the end of Silverman IX (of course, these computations are, as you say, not for the faint hearted) – Mummy the turkey Feb 15 '21 at 00:14